Suppose a sequence of derivatives of functions $f'_n $ converge uniformly to $f'$ where $f_n$ is defined on the on the interval $[a,b]$. And $f_n(x)$ converges pointwise to $f(x)$ for $x\in \mathbb{R}$. Prove that $f_n$ converges uniformly to $f$.

$2.$ Another point to note may be is : there exists a point $c$ such that $$\frac{f'_n(c)}{f'(c)}=\frac{f_n(b)-f_n(a)}{f(b)-f(a)}$$
Applying limit to both side$$\lim_{n\rightarrow\infty}\frac{f_n(b)-f_n(a)}{f(b)-f(a)}=1$$

It seems second one is right track but I don't seem to be able to complete it.

Approach 2 might go somewhere if you set $a=x_0$ and $b=x_0+h$.Then you get that $f_n(x_0+h)-f_n(x_0)\to f(x_0+h)-f(x_0)$ uniformly with respect to $h$. Since $f_n(x_0)\to f(x_0)$, you might conclude from this that $f_n(x_0+h)\to f(x_0+h)$ uniformly in $h$, which is the same thing as to say that $f_n(x)\to f(x)$ uniformly in $x$.
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Giuseppe NegroJan 27 '13 at 17:40

Thank You. This was one I was looking for. The integration one is good too.
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user45099Jan 27 '13 at 17:46

The only small disadvantage of this proof is that it requires the fundamental theorem of calculus and so it requires integrability of $f'_n$. This is certainly the case if $f_n\in C^1([a,b])$, but the theorem still holds for more pathological functions which are not the integral of their derivative. For this (small) generalization you can consult Rudin, Principles of mathematical analysis, chapter 7.